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This is the full question:

$f: S \to T$ is a function, then the following statements are equivalent:

a) $f$ is one-one on $S$

b) $f(A \cap B) = f(A)\cap f(B) \, \forall \, A,B \subseteq S $

c) $f^{-1}(f(A))=A \hspace{4mm} \forall \, A\subseteq S $

d) $f(A) \cap f(B) = \emptyset \, \forall \, A,B \subseteq S, $ and $ A \cap B \, = \emptyset$

e) $f(A-B)=f(A)-f(B) \,\, \forall \, A,B \subseteq S$ and $B \subseteq A$

I am having trouble is understanding a step in the proof of $\, d \implies e$ , Here is the proof

If $B \subseteq A \text{ then } (A-B)\cap B = \emptyset \\ \implies f(A-B) \cap f(B) = \emptyset \\ \implies f(A-B) \subseteq f(A)-f(B) \hspace{5mm} (*) \\ \text{if } \, y \in f(A-B) \text{ then } y=f(x) \text{ for some } x \in A \text{ and } x \notin B \\ x\in A-B \implies f(x)=y \in f(A-B) \\ \implies f(A)-f(B) \subseteq f(A-B) $

How did we get statement (*)? I didn't understand that step.

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  • $\begingroup$ d) is hard to parse logically. Do you want to say "$f(A)\cap f(B)=\emptyset$ for all $A,B\subseteq S$ with $A\cap B=\emptyset$"? (Similar for e) $\endgroup$ Commented Aug 23, 2019 at 7:40
  • $\begingroup$ yes, that is what i meant @HagenvonEitzen $\endgroup$
    – Sam
    Commented Aug 23, 2019 at 8:33

3 Answers 3

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Let $y \in f (A-B)$. Then certainly $y \in f(A)$. To prove (*) we have to show that $y \notin f(B)$. Prove by contradiction. Suppose, if possible, $y \in f(B)$. Then $y \in f(A-B)\cap f(B)$ but this intersection is empty.

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Well, if $B\subseteq A$ and $f(A-B)\cap f(B)=\emptyset$, then $f(A-B)\subseteq f(A)-f(B)$, since $f(A-B)\subseteq f(A)$.

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We know that $f(A - B) \subset f(A)$. Then, it must also be that, as $f(B) \subset f(A)$, because $f(A) = f(A - B) \cup f(B)$. Thus, we know that the set of $f(A-B)$ must be contained in $f(A)$.

However, the statement preceding (*) tells us that $f(A - B) \cup f(B) = \emptyset$. Or, in other words, it tells us that $f(A - B)$, which is contained in $f(A)$, does not intersect $f(B)$ at all. Thus, by definition, $f(A - B)$ is contained within (and is thus a subset of) that subset of $f(A)$ which contains no points from $f(B)$. This subset of $f(A)$ is, by definition, $f(A) - f(B)$.

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